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What is the mass of the planet?

  1. Jun 20, 2007 #1
    I have a few questions that I'm not sure on how to do:

    1) A 20kg mass is connected to a 8kg mass by a thin frictionless string on a pulley. It accelerates to 1.5m/s2 and the heavier mass slides along a rough surface. What is the coefficient of friction?

    3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?

    4) A 7.5x10^3kg sattelite orbits a planet at an altitude of 300km. The radius of the planet is 8.5x10^6. The gravitational field at the radius is 6.5N/kg. What is the mass of the planet?

    5)A rocket car uniformly accelterates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?

    Need help with these questions ASAP.. Thanks.
     
  2. jcsd
  3. Jun 20, 2007 #2
    I am so tired of people posting threads like this. You have obviously not read the guidelines regarding the posting of homework questions.

     
  4. Jun 20, 2007 #3
    Sorry I have a test tomorrow and am lost on how to do these questions.. Just need some help on how to get started..
     
  5. Jun 20, 2007 #4
    I will certainly try to help you, but you have to show me that you've at least thought about these questions prior to asking for help. We're here to help you learn, not do your homework for you.

     
  6. Jun 20, 2007 #5
    2) 3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?
    For this one would I use E=mc^2 and just fill it in with the data provided?

    5) A rocket car uniformly accelerates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?

    For this one I think I have the answer. We know:
    X:
    a= 15m/s2
    d=?
    vi= 0m/s

    Y:
    d= -75m
    a= -9.8m/s2
    vi= 0.0m/s
    t=?

    Y: d= vit+.5at^2
    and when solving for t, I got 2.8s.

    then for X: d= vit+.5at^2
    a= 15m/s2
    d=?
    vi= 0m/s
    t= 2.8s

    Answer: d= 59m

    As for the other two, I do not know what to do for them as #1 I cannot find examples for and #4 I haven't done before so I'm completely lost on those and these are the ones that will be on the exam, so I just need something to get me started on them, thanks.
     
  7. Jun 21, 2007 #6
    Anyone able to help me out?
     
  8. Jun 21, 2007 #7

    nrqed

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    Yes. Make sure you put the mass in kilograms
    This question does not make sense to me. If we don't know for how long it accelerates before reaching the cliff, we don't know its horizontal speed when it goes over the cliff and it is impossible to find how far it will land.
     
  9. Jun 21, 2007 #8
    *The distance from the car to the edge of the cliff is 100m.
     
    Last edited: Jun 21, 2007
  10. Jun 21, 2007 #9

    nrqed

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    :eek: Ah!! It was not in the original question.

    Ok, but then your solution is wrong. You did not use hat information. And what you did is to use the time of fall in the vertical direction (2.8 sec) as the time for the horziontal motion. There is no reason to assume that those two times are equal!

    What you need to do:

    a) find the horizontal velocity when the car reaches the cliff using 100 m and the 15 m/s^2 (are you sure it's 15!?! This is too large for a real car)

    b) find the time it takes to fall 75 m. That you already did. I did not check but if you did it correctly, it will be 2.8 s (you might want to keep more sig figs)

    c) finally, use the x velcoity of part a and the time of part b to find the horizontal distance travelled during the fall
     
  11. Jun 21, 2007 #10

    nrqed

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    The equations are

    Along x

    Eq 1: [itex] x_f = x_i + v_{xi} t + 1/2 a_x t^2 [/itex]

    Eq 2: [itex] v_{xf} = v_{xi} + a_x t [/itex]


    From the previous two, one can also show that

    Eq 3: [itex] v_{xf}^2 = v_{xi}^2 + 2 a_x (x_f-x_i) [/itex]

    Along y (assuming only the force of gravity is present)

    Eq 4 [itex] y_y = y_i + v_{yi} t - 1/2 g t^2 [/itex]

    Eq 5 [itex] v_{yf} = v_{yi} - g t [/itex]

    Eq 3: [itex] v_{yf}^2 = v_{yi}^2 - 2 g (y_f-y_i) [/itex]
    where I use g = +9.8 m/s^2 (my g is positive).


    Along x, if you now that the distance (x_f -x_i) is 100 m and you know a_x, and v_xi=0, use eq 4 to find the value of v_xf when the car reaches the cliff



    Hope this helps
     
  12. Jun 21, 2007 #11
    I got 54.8m/s as the velocity and before you said that the x and y time are not the same so should I use the y time to solve for the distance? Using the y time I get 157.4 as the ditance. Nvm got diff answer and gotta go now, thanks for trying to help me though.
     
    Last edited: Jun 21, 2007
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